Angular momentum of the electron in a hydrogen atom is 3h/2π, then find the energy of the electron in that orbit:

The correct option is −1.51eV.

To find the energy of the electron in the given orbit, we can follow these steps:

Step 1: Determine the principal quantum number (n) using Bohr's quantization condition.
According to Bohr's postulates for the hydrogen atom, the angular momentum ($L$) of an electron in a stable orbit is quantized and given by:
$L=\frac{nh}{2\pi }$
where:
• $n$ is the principal quantum number (orbit number)
• $h$ is Planck's constant

Step 2: Match the given angular momentum value.
We are given that the angular momentum of the electron is:
$L=\frac{3h}{2\pi }$
By comparing this to the quantization formula:
$\frac{nh}{2\pi }=\frac{3h}{2\pi }$
Thus, the principal quantum number is:
$n=3$

Step 3: Calculate the energy of the electron in the third orbit (n = 3).
The energy ($E_n$) of an electron in the $n$^th orbit of a hydrogen atom is given by:
$E_n=-\frac{13.6}{n^2}\text{eV}$
Substituting $n=3$ into this relation:
$E_3=-\frac{13.6}{3^2}\text{eV}$
$E_3=-\frac{13.6}{9}\text{eV}$
$E_3\approx -1.51\text{eV}$

Therefore, the energy of the electron in that orbit is −1.51eV.